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Showing that $S^{-1}$ is continuous is the same as showing that $S$ is a closed map. For this, we can use the oft-cited result from general topology:

Let $f\colon X\to Y$ be a continuous bijection from a compact space $X$ to a Hausdorff space $Y$. Then $f$ is closed, and hence a homeomorphism.

The proof of this is simple. If $E\subseteq X$ is closed, then $E$ is compact since $X$ is compact. Since $f$ is continuous, it follows that $f(E)$ is compact. Since compact subsets of Hausdorff spaces are closed, it follows that $f(E)$ is closed.